\newproblem{lay:4_3_24}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 4.3.24}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $B=\{\mathbf{v}_1,\mathbf{v}_2,...,\mathbf{v}_n\}$ be a linearly independent set of $\mathbf{R}^n$. Explain why $B$ must be basis for $\mathbb{R}^n$.
}{
  % Solution
	To be a basis for $\mathbb{R}^n$ a set needs to be linearly independent ($B$ is so by hypothesis) and span $\mathbb{R}^n$. Let's check this latter requirement.
	Let us form the matrix $A=\begin{pmatrix}\mathbf{v}_1 & \mathbf{v}_2 & ... &\mathbf{v}_n\end{pmatrix}$. $B$ spans $\mathbb{R}^n$ if for any vector
	$\mathbf{b}\in\mathbb{R}^n$, the matrix equation $A\mathbf{x}=\mathbf{b}$ has a solution. Since the columns of $A$ are linearly independent, by the invertible
	theorem matrix (Theorem 5.1, Chapter 3), we know that the matrix equation above has a solution, and consequently, the columns of $A$ (that is $B$) spans $\mathbb{R}^n$.
}
\useproblem{lay:4_3_24}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
